Question

If \[{}^{n}{{C}_{n-2}}=15\]then find n.

Answer 1

Hint: To solve this question at first we have to expand \[{}^{n}{{C}_{n-2}}\] using a combination formula. Then we solve for the expression in terms of n and equate the expression with 15 to get an equation. By solving the obtained equation we will get the value of n.

Complete step-by-step answer:
From the definition of combination we know the formula,
\[{}^{n}{{C}_{r}}=\dfrac{n!}{(n-r)!r!}\] , .………………. (1), which shows number of ways to choose ‘r’ items from total ‘n’ items
By replacing ‘r’ with \[\left( n-2 \right)\]in eq. (1), we will get
\[{}^{n}{{C}_{n-2}}=\dfrac{n!}{\left\{ n-\left( n-2 \right) \right\}!\left( n-2 \right)!}\]
On solving bracket by adding n and - n, we get
\[{}^{n}{{C}_{n-2}}=\dfrac{n!}{2!\left( n-2 \right)!}\]……………………………………………… (2)
We now that the factorial of a positive integer n is given by,
\[n!=n\left( n-1 \right)\left( n-2 \right)\times ........\times 3\times 2\times 1\] ……………………………. (3)
For example,
\[6!=6\times 5\times 4\times 3\times 2\times 1=720\]
In a similar manner,
\[\left( n-2 \right)!=\left( n-2 \right)\left( n-3 \right)\times ............\times 3\times 2\times 1\] ……………………… (4)
Now substituting the values of eq. (3) and (4) in eq. (2), we will get
\[\Rightarrow {}^{n}{{C}_{n-2}}=\dfrac{n\left( n-1 \right)\left( n-2 \right)\times ........\times 3\times 2\times 1}{2!\times \left( n-2 \right)\left( n-3 \right)\times ............\times 3\times 2\times 1}\]
Eliminating all the common terms from denominator and numerator by cancelling them out, we get
\[\Rightarrow {}^{n}{{C}_{n-2}}=\dfrac{n\left( n-1 \right)}{2}\]………….. (5)
Here we got the expression for\[{}^{n}{{C}_{n-2}}\]. To verify the expression let’s take some examples.
For\[n=3\],
\[\begin{align}
  & {}^{3}{{C}_{3-2}}={}^{3}{{C}_{1}}=\dfrac{3!}{\left\{ 3-1 \right\}!1!}=\dfrac{3\times 2\times 1}{2\times 1\times 1}=3 \\
 & \dfrac{n\left( n-1 \right)}{2}=\dfrac{3\times 2}{2}=1 \\
\end{align}\]
For\[n=5\],
\[\begin{align}
  & {}^{5}{{C}_{5-2}}={}^{5}{{C}_{3}}=\dfrac{5!}{\left\{ 5-3 \right\}!3!}=\dfrac{5\times 4\times 3\times 2\times 1}{2\times 1\times 3\times 2\times 1}=10 \\
 & \dfrac{n\left( n-1 \right)}{2}=\dfrac{5\times 4}{2}=10 \\
\end{align}\]
But according to question,
\[{}^{n}{{C}_{n-2}}=15\] ............................................................ (6)
Now comparing eq. (5) and (6) we will get,
\[\begin{align}
  & \Rightarrow \dfrac{n\left( n-1 \right)}{2}=15 \\
 & \Rightarrow n\left( n-1 \right)=30 \\
 & \Rightarrow {{n}^{2}}-n-30=0 \\
\end{align}\]
Factorizing, the quadratic equation by splitting, - n as -6n+5n, we get
\[\begin{align}
  & \Rightarrow {{n}^{2}}-6n+5n-30=0 \\
 & \Rightarrow n(n-6)+5(n-6)=0 \\
\end{align}\]
Taking, ( n -6 ) common from equation, we get
\[\Rightarrow (n-6)(n+5)=0\]
On comparing Left hand side and right hand side, we get
\[\Rightarrow n=6,-5\]…………………………………………. (7)
But we know that n must be a positive integer from the definition of combination.
Hence the required value of n is 6.

Note: Factorial of a number n that is \[n!\] defined only for any nonnegative integer n. Therefore here we cannot take the value for\[n=-5\]. Also, use can solve the quadratic equation $a{{x}^{2}}+bx+c=0$ to get value of x, by using quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . Try not to make any calculation errors while solving the question, as it will change the final answer.